Theorem sbalex 2242

Description: Equivalence of two ways to express proper substitution of a setvar for another setvar disjoint from it in a formula. This proof of their equivalence does not use df-sb 2073.

That both sides of the biconditional express proper substitution is proved by sb5 2274 and sb6 2093. The implication "to the left" is equs4v 2009 and does not require ax-10 2143 nor ax-12 2177. It also holds without disjoint variable condition if we allow more axioms (see equs4 2415). Theorem 6.2 of [Quine] p. 40. Theorem equs5 2459 replaces the disjoint variable condition with a distinctor antecedent. Theorem equs45f 2458 replaces the disjoint variable condition on 𝑥, 𝑡 with the nonfreeness hypothesis of 𝑡 in 𝜑. (Contributed by NM, 14-Apr-2008.) Revised to use equsexv 2267 in place of equsex 2417 in order to remove dependency on ax-13 2371. (Revised by BJ, 20-Dec-2020.) Revise to remove dependency on df-sb 2073. (Revised by BJ, 21-Sep-2024.)

Assertion

Ref	Expression
sbalex	⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑))

Distinct variable group:   𝑥,𝑡

Allowed substitution hints:   𝜑(𝑥,𝑡)

Proof of Theorem sbalex

Step	Hyp	Ref	Expression
1		nfa1 2154	. . 3 ⊢ Ⅎ𝑥∀𝑥(𝑥 = 𝑡 → 𝜑)
2		ax12v2 2179	. . . 4 ⊢ (𝑥 = 𝑡 → (𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜑)))
3	2	imp 410	. . 3 ⊢ ((𝑥 = 𝑡 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑡 → 𝜑))
4	1, 3	exlimi 2217	. 2 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑡 → 𝜑))
5		equs4v 2009	. 2 ⊢ (∀𝑥(𝑥 = 𝑡 → 𝜑) → ∃𝑥(𝑥 = 𝑡 ∧ 𝜑))
6	4, 5	impbii 212	1 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399  ∀wal 1541  ∃wex 1787

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2018  ax-10 2143  ax-12 2177

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-ex 1788  df-nf 1792

This theorem is referenced by:  equsexv  2267  sb5  2274  dfsb7  2282  mopick  2626  alexeqg  3548  dfdif3  4015  pm13.196a  41646